Question

Elizabeth lives in San Francisco and works in Mountain View. In the morning, she has 333 transportation options (take a bus, a cab, or a train) to work, and in the evening she has the same 333 choices for her trip home. If Elizabeth randomly chooses her ride in the morning and in the evening, what is the probability that she'll use a cab exactly one time?

Answer 1

4/9

Explanation:

Answer 2

All transportation (bus, cab, train) are all similarly likely to be selected, and 1 of them must be selected at morning and evening, so we get: P (bus) = P (cab) = P (train) = 1/3. We also have P(no cab in evening) = P(no cab at morning) = 2/3
Now, P(using cab exactly once) = P(cab at morning and no cab in the evening) + P(no cab at morning and cab in the evening)
= P(cab, no cab) + P(no cab, cab)
= 1/3 * 2/3 + 2/3 * 1/3
= 2/9 + 2/9
= 4/9
Probability that Elizabeth uses a cab only once is 4/9.